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이 문서에서는 PHP 중국어 웹사이트 매뉴얼 풀어 주다
The mysqli extension features a dual interface. It supports the procedural and object-oriented programming paradigm.
Users migrating from the old mysql extension may prefer the procedural interface. The procedural interface is similar to that of the old mysql extension. In many cases, the function names differ only by prefix. Some mysqli functions take a connection handle as their first argument, whereas matching functions in the old mysql interface take it as an optional last argument.
Example #1 Easy migration from the old mysql extension
<?php
$mysqli = mysqli_connect ( "example.com" , "user" , "password" , "database" );
$res = mysqli_query ( $mysqli , "SELECT 'Please, do not use ' AS _msg FROM DUAL" );
$row = mysqli_fetch_assoc ( $res );
echo $row [ '_msg' ];
$mysql = mysql_connect ( "example.com" , "user" , "password" );
mysql_select_db ( "test" );
$res = mysql_query ( "SELECT 'the mysql extension for new developments.' AS _msg FROM DUAL" , $mysql );
$row = mysql_fetch_assoc ( $res );
echo $row [ '_msg' ];
?>
以上例程会输出:
Please, do not use the mysql extension for new developments.
The object-oriented interface
In addition to the classical procedural interface, users can choose to use the object-oriented interface. The documentation is organized using the object-oriented interface. The object-oriented interface shows functions grouped by their purpose, making it easier to get started. The reference section gives examples for both syntax variants.
There are no significant performance differences between the two interfaces. Users can base their choice on personal preference.
Example #2 Object-oriented and procedural interface
<?php
$mysqli = mysqli_connect ( "example.com" , "user" , "password" , "database" );
if ( mysqli_connect_errno ( $mysqli )) {
echo "Failed to connect to MySQL: " . mysqli_connect_error ();
}
$res = mysqli_query ( $mysqli , "SELECT 'A world full of ' AS _msg FROM DUAL" );
$row = mysqli_fetch_assoc ( $res );
echo $row [ '_msg' ];
$mysqli = new mysqli ( "example.com" , "user" , "password" , "database" );
if ( $mysqli -> connect_errno ) {
echo "Failed to connect to MySQL: " . $mysqli -> connect_error ;
}
$res = $mysqli -> query ( "SELECT 'choices to please everybody.' AS _msg FROM DUAL" );
$row = $res -> fetch_assoc ();
echo $row [ '_msg' ];
?>
以上例程会输出:
A world full of choices to please everybody.
The object oriented interface is used for the quickstart because the reference section is organized that way.
Mixing styles
It is possible to switch between styles at any time. Mixing both styles is not recommended for code clarity and coding style reasons.
Example #3 Bad coding style
<?php
$mysqli = new mysqli ( "example.com" , "user" , "password" , "database" );
if ( $mysqli -> connect_errno ) {
echo "Failed to connect to MySQL: " . $mysqli -> connect_error ;
}
$res = mysqli_query ( $mysqli , "SELECT 'Possible but bad style.' AS _msg FROM DUAL" );
if (! $res ) {
echo "Failed to run query: (" . $mysqli -> errno . ") " . $mysqli -> error ;
}
if ( $row = $res -> fetch_assoc ()) {
echo $row [ '_msg' ];
}
?>
以上例程会输出:
Possible but bad style.
See also
[#1] Anonymous [2015-01-06 23:36:58]
Just want to add that both procedural mysqli_connect_errno and mysqli_connect_error DON'T accept any arguments!
http://php.net/manual/de/mysqli.connect-errno.php
http://php.net/manual/de/mysqli.connect-error.php
"int mysqli_connect_errno ( void )"
"string mysqli_connect_error ( void )"
It clearly states "void" there.
Adding the mysqli-Instance as a parameter makes it look like it pulls the error-number out of the provided instance, which is not actually happening. This could end in a hard to detect bug when connecting to multiple SQL servers.
And it is confusing for beginners.
[#2] Anonymous [2013-07-09 17:26:57]
After significant searching with no success finding a solution I was able to figure out the following to error 2005 where the connection could not find the host. While I am hosted on 1&1 I saw numerous posts with similar problems.
The solution turned out to be as follows:
<?php
$mysqli = new mysqli(DB_HOST, DB_USER, DB_PASS, DB_NAME, DB_PORT, DB_UNIX_SOCKET);
I added the DB_PORT and DB_UNIX_SOCKET
define("DB_NAME", "YOUR DB NAME");
define("DB_USER", "YOUR DB USER");
define("DB_PASS", "YOUR DB PWORD");
define("DB_UNIX_SOCKET", "/tmp/mysql5.sock");
define("DB_PORT", 3306);
?>